PS2_sol(4).pdf - Problem Set 2 Conditional Expectations and...

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Problem Set 2: Conditional Expectations and Prediction ECON 21200, Spring 2015 SOLUTIONS Problem 1 (AR(2) Dynamics) a) For each AR(2) process listed below, calculate the roots of the lag polyno- mial. Based on the roots, determine if the process is stationary. 1. Y t = 1 . 4 Y t - 1 - 0 . 48 Y t - 2 2. Y t = Y t - 1 + 1 . 2 Y t - 2 3. Y t = 1 . 6 Y t - 1 - 0 . 8 Y t - 2 4. Y t = Y t - 1 - 1 . 2 Y t - 2 Solution: For each process, calculate the roots of the characteristic polynomial. The process in general is Y t = φ 1 Y t - 1 + φ 2 Y t - 2 so the characteristic polynomial is λ 2 - φ 1 λ - φ 2 . The roots for each process are: 1. ( λ 1 , λ 2 ) = (0 . 8 , 0 . 6) 2. ( λ 1 , λ 2 ) = (1 . 7 , - . 7) 3. ( λ 1 , λ 2 ) = (0 . 8 + 0 . 4 i, 0 . 8 - 0 . 4 i ) 4. ( λ 1 , λ 2 ) = (0 . 5 + 0 . 97 i, 0 . 5 - 0 . 97 i ) We can see immediately the first is stationary, and the second is not. Calculate the modulus of the complex numbers to see that the third is stationary (modulus=0.89) and the fourth is not (modulus=1.095). 1
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b) For each process in part (a), convert the roots into polar coordinates, λ = Ce where ω ( - π, π ] and C 0 . For the first process, the roots are already in polar coordinates, where ω = 0. For the second process, the positive root is already in polar coordinates; the second would be written as 0 . 7 e . For the last two processes, since you already calculate the modulus, you just need to find ω . If we write λ = a + bi then ω = arctan b a . For the third process, we get λ 1 = 0 . 89 exp(0 . 46 i ), λ 2 = 0 . 89 exp( - 0 . 46 i ). For the fourth process, we get λ 1 = 1 . 095 exp(1 . 09 i ), λ 2 = 1 . 095 exp( - 1 . 09 i ). c) The roots of the characteristic polynomial imply something about the cycli- cality of the process. In particular, a root Ce implies a periodicity of 2 π | ω | . What are the periodicities of the roots for each process in part (a)? The periodicity of the first process is infinite (that is, there are no cycles at all). For the second process, we expect a cycle of 2 π π = 2. For the third process, the periodicity is 2 π 0 . 46 = 13 . 55. For the fourth process, the periodicity is 2 π 1 . 09 = 5 . 76. d) Compute the impulse response function for each process in part (a) by setting Y t = 0 for t < 0 , 0 = 1 and t = 0 for t > 0 . Then calculate Y 0 , Y 1 , . . . , Y 50 . Instead of reporting in a table, plot these values as a function of time. Solution: See Figure 1. e) For each impulse response function, comment on the patterns you see. Which impulse response functions are cyclical? What are the periods of the cycles? Which IRF’s converge? Which diverge? How are these patterns re- lated to the roots of the characteristic polynomial? Solution: We can see that the non-stationary processes have impulse response functions that diverge, while the stationary processes have IRF’s that converge. Second, the processes with real roots have monotonic IRF’s. Actually, the second process is supposed to have a cycle of length 2 but that is swamped by the fact that the non-stationary root is positive and causes the IRF to diverge.
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